What is the back emf of a \(120V\) motor that draws \(8.00A\) at its normal speed and \(20.0A\) when first starting?

Current is the term used to describe the speed at which charge moves. Resistance is the propensity of a substance to oppose the flow of charge.

\(\begin{align}{}{I_{start{\rm{ }}}} = 20\;A\\{I_{back{\rm{ }}}} = 8\;A\\V = 120\;V\end{align}\)

The current in a circuit,

\(I = \frac{V}{R},\)

where \(V\) is the potential and \(R\) is the resistance.

First we need find the resistance value,

\(\begin{align}{}I = \frac{V}{R}\\{I_{start{\rm{ }}}} = \frac{V}{R}\\20 = \frac{{120}}{R}\\R = 6\Omega \end{align}\)

Now finding the back emf

\(\begin{align}{}{I_{back{\rm{ }}}} = \frac{{V - {\rm{ }}Back{\rm{ }}emf{\rm{ }}}}{R}\\8 = \frac{{120 - {\rm{ }}Back{\rm{ }}emf{\rm{ }}}}{6}\\Backemf = 72\;V\end{align}\)

Therefore, the emf is obtained as \(72\;V\).